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Operations Research by

FUAD MUSTEFA

Linear Programming

CHAPTER TWO

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Introduction to Linear Programming

Linear Programming is that branch of mathematical programming which is designed to solve optimization problems where all the constraints as will as the objectives are expressed as Linear function .

Linear Programming is a technique for making decisions under certainty i.e.; when all the courses of actions available to an organization are known & the objective of the firm along with its constraints are quantified.

LP is the analysis of problems in which a Linear function of a number of variables is to be optimized (maximized or minimized) when whose variables are subject to a number of constraints in the mathematical inequalities.

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Intro cont…

From the above definitions, it is clear that:

LP is an optimization technique

It deals with the problem of allocation of scarce sources

It generates solutions based on the feature and characteristics of the actual problem or situation.

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Intro cont…

Product-mix

problems

Investment planning problems

Blending strategy formulations and

Marketing &

Distribution

management.

LP has been highly successful in solving

the following types of problems :

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Intro cont…

All LPP have the following properties in common .

• The objective is always the same• Presence of constraints which limit the extent to which the

objective can be pursued/achieved.• Availability of alternatives• The objectives and constraints can be expressed in the form of

linear relation.

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Basic Assumptions of Linear Programming

Decision or Activity Variables & Their Inter-Relationship

Finite Objective Functions

Limited Factors/Constraints

Presence of Different Alternatives

Non-Negativity Restrictions

Linearity Criterion

Additive

Mutually Exclusive Criterion

Divisibility

Certainty

Finiteness.

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Application Areas of Linear Programming

Military Applications

Agriculture

 Environmental Protection.

Facilities Location

Product-Mix

Production

Mixing or Blending

Transportation & Trans-Shipment

Portfolio Selection

Profit Planning & Contract.

Traveling Salesmen Problem

Media Selection/Evaluation

Staffing

Job Analysis

Wages and Salary Administration.

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LPP Formulation

Steps in Formulating a LP ModelStep I Identification of the decision variables

Step II Identification of the constraints

Step III formulate the objective function

Step IV formulate as LPP

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Example 3F furniture Ltd. manufactures two products, tables & chair. Both the products have to be processed through two machines Ml & M2 the total machine-hours available are: 200 hours of M1 and 400 hours of M2 respectively. Time in hours required for producing a chair and a table on both the machines is as follows:

Machine Table Chair

M1 7 4

M2 5 5

Profit from the Sale of table is Birr 40 and that of a chair is Birr 30.Required: formulate the LP Problem (LPP)?

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Example Step I Identification of the decision variables. Let x1 = Number of

tables produced and X2 = Number of Chairs produced

Step II List down all the constraints :A. Total time on machine M1 cannot exceed 200 hour ∴ 7X1+4X2≤200 and

B. Total time on machine M2 cannot exceed 400 hour ∴ 5X1+5X2≤400Step III. The objective function for maximizing the profit is given by.Minimize Z= 40X1+30X2

Presenting the problem as LPPMinimize Z= 40X1+30X2 subject to: 7X1+4X2≤200 5X1+5X2≤400 X1&X2≥0

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Example 2

Alpha Limited company produces & sells 2 different products under the brand name black & white. The profit per unit on these products is Birr 50 & Birr 40 respectively. Both black & white employ the same manufacturing process which has a fixed total capacity of 50,000 man-hour. As per the estimates of the marketing research department of Alpha Limited, there is a market demand for maximum 8,000 units of Black & 10,000 units of white. Subject to the overall demand, the products can be sold in any possible combination. If it takes 3 hours to produce one unit of black & 2 hours to produce one unit of white. Formulate the above problem as a linear programming model?

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Example 3

3. An agriculturalist has a 125 acre farm. He produce onion, tomato and potato. What ever he rises is fully sold in the market. He gets Birr 5 for onion per kg, Birr 4 for tomato per kg and Birr 5 for potato per kg. The average per acre yield is 1500kg of onion, 1800kg of tomato and 1200kg of potato. To produce each 100kg of onion and tomato and 80kg potato, a sum of Birr 12.5 has to be used for fertilizer. Labor required for each acre to rise the crop is 6 man days for onion and potato each and 5 man-days for tomato. A total of 500 man-days of labor at a rate of Birr 40 per man-day are available. Formulate this as LP model to maximize the agriculturist’s total profit.

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Method for solving LPP

There are two types of finding a solution for Linear programming

problems.

• Graphic solution and• Simplex solution

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Graphical Method of solving Linear Programming Problems

The graphic solution procedure is one of the methods of solving two variable linear programming problems.

Steps in graphic solution method

Step I Defining the problem

Step II Plot the constraints Graphically

Step III Locate the solution space: Solution space or the feasible region is the graphical area which satisfies all the constraints at the same time. Such a solution point (x, y) always occurs at the corner Points of the feasible Region the feasible region is determined as follows:

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Graphic solution cont…

1. For "greater than" & "greater than or equal to" constraints, the feasible region or the solution space is the area that lays above the constraint lines.

2. For" Less Then" &" Less than or equal to" constraint, the feasible region or the solution space is the area that lays below the constraint lines.

Step IV Selecting the graphic solution technique

There are two graphic techniques to find solution;

a. Corner Point Method and

b. Iso-profit (or Iso-cost) method may be used

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Corner Point Method to find Solution for Graphics Method

Since the solution point (x, y) always occurs at the corner point of the feasible or solution space, identify each of the extreme points or corner points of the feasible region by the method of simultaneous equations.

By putting the value of the corner point's co-ordinates [e.g. (2, 3)] into the objective function, calculate the profit (or the cost) at each of the corner points.

In a Maximize problem, the optimal solution occurs at that corner point which gives the highest profit. In a minimization problem, the optimal solution occurs at that corner point which gives the lowest profit.

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Graphic solution Example 1

XYZ Ltd. Co. Wishes to purchase a maximum of 3600 units of two types of product, A & B are available in the market. Product A occupies a space of 3 cubic feet & cost Birr 9 whereas B occupies a space of 1 cubic feet & cost Birr 13 per unit. The budgetary constraints of the company do not allow spending more than Birr 39,000. The total availability of space in the company's god own is 6000 cubic feet. Profit margin of both the product A & B is Birr 3 & Birr 4 respectively. Formulate the above problem as a linear programming model and solve it by using graphical method. You are required to ascertain the best possible combination of purchase of A & B so that the total profits are maximized.

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Graphic solution EG. Cont…

Number of unites o A

Y

C

6000

3600

3000

2000 3600 3900/9X

Num

ber

of u

nite

s of

B

A1

(0, 0)

B(1300,2100)

3X+Y≤6000 (Storage area constraint)

X+Y≤3600(Maximum unites constraint)

9X+13Y≤3900(Budgetary constraint)

o

Feasible region

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At corner points (O, A, B, C), find the profit value from the objective function. Those points which maximize the profit are the optimal

point.

Corner point Coordinates Objective function Value

O (0,0) Z=0+0 0

A (0,3000) Z=0+4x3000 12,000

B (1300,2100) Z=3x1300 + 4 x2100 12,300

C (2000,0) Z=3 x 2000 + 0 6000

20

2. Suppose that a machine shop has two different types of machines; machine 1 and machine 2, which can be used to make a single product .These machine vary in the amount of product produced per hour, in the amount of labor used and in the cost of operation.

Assume that at least a certain amount of product must be produced and we would like to utilize at least the regular labor force. How much should we utilize in each machine in order to utilize total costs and still meets the requirement? The resource used, the cost and the required hour is given in the following table.

Machine 1

(X)

Machine 2

(Y)

Minimum required hours

Product produced/hr 20 15 100

Labor/hr 2 3 15

Operation Cost Birr25 Birr30

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B (2.5, 3.33)

A (0, 20/3)

C (7.5, 0)

Y

5

X2 =0

Y=0

Feasible Region

2X1 + 3X2 ≥ 15

20X+15Y≥100

5

X

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Corner point Coordinates Objective function Value

A

B

C

(0,20/3)

(2.5, 3.33)

(7.5, 0)

Z=0+20/3×30

Z=2.5×25+3.33x30

Z=7.5x25+0

200

162.5

187.5

21 3025. XXZMinimize

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Special Cases in Graphics Methods

The following are the major special cases in graphics solution

a. Alternative Optima

b. Infeasible Solution

c. Unbounded solution

a. Alternative Optimal solutionWhen the objective function is parallel to a binding constraint; (a constraint that is satisfied in the equality sense by the optimal solution), the objective function will assume the same optimal value at more than one solution point, for this reason they are called alternative optima.

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Example :The information given below is for the products A and B.

Machine hours per week Maximum availableDepartment Product A Product B per weekCutting 3 6 900Assembly 1 1 200Profit per unit Birr 8 Birr 16

Assume that the company has a marketing constraint on selling products B and therefore it can sale a maximum of 125 units of this product.Required:a. Formulate the LPP of this problem?b. Find the optimal solution?

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The LPP Model of the problem is:

0,

125

200

90063

:

168.

21

2

21

21

21

XX

X

XX

XX

Subjectto

XXZMax

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D (100,100)

(0, 200) (0,150)

B (0, 125)

X2

A (0, 0)

C (50, 125)

X1=0

Cutting: 3X1+6X2=900FR

X1

X2=125 Marketing equation

(300, 0) (200, 0)

20021 XX

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Infeasible Solution

Sometimes possible that the constraints may be inconsistent so that there is no feasible solution to the problem. Such a situation is called infeasibility.

Example:Maximize Z=20X1+30X2

Subject to: 2X1+X2< 40

4X1+X2< 60

X1 > 30 and X1, X2 > 0

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X1

X2

X1=0

X2=0

(0, 60)

(0, 40)

X1=30

2X1+X2= 40

4X1+X2= 60

(15, 0) (20, 0) (30, 0)

Note: -In the ff graph, there is no common point in the shaded area.

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Unbounded Solution

When the value of decision variables in LP is permitted to increase infinitely without violating the feasibility condition, then the solution is said to be unbounded

Example:Use the graphical method to solve the following LPP.1. Maximize Z=3X1+4X2

Subject to:X1-X2<-1==> -X1+X2>1 since the quantity solution is positive

-X1+X2<0

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Unbounded Feasible region

Feasible Region

X1+X2

=0

X1-X2 =-1

X1

X2

1

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Maximize Z=3X1+2X2

Subject to: X1-X2<1

X1+X2≥3

X1

X2

Unbounded

Feasible Region A (0, 3)

B (2, 1)X1+X2

=3

X1-X2=1

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Simplex Method

The Simplex method is an iterative or “step by step” method or repetitive algebraic approach that moves automatically from one basic feasible solution to another basic feasible solution improving the situation each time until the optimal solution is reached at.

The Simplex method starts with a corner that is in the solution space or feasible region and moves to another corner, the solution space improving the value of the objective function each time until optimal solution is reached at the optimal corner.

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Standard forms of (LP) problems

1. All the constraints should be expressed as equations by adding slack or surplus and/or artificial variables.

2. The right hand side of each constraint should be made non-negative; if it is not, this should be done by multiplying both side of the resulting constraint by -1.

3. The objective function should be of the maximization type.

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The standard form of LP problem is expressed as follow;Optimize (Maximize or Minimize) Z= Subject to the linear constraint:

. . . . . .

And

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Remarks; three types of additional variables, namely•Slack variable (s)•Surplus variable (-s), and •Artificial variable (A)Are added in the given LP problem to convert it in to the standard form for the following reasons: •These variables allow as converting inequalities in to equalities, there by converting the LP problem in to a form that is amenable to algebraic solution. •These variables permit as to make a more comprehensive economic interpretation of a final solution.•Help us to get an initial feasible solution presented by the column of the identity matrix.

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The summery of the extra variable to be added in the given LP problem to convert it in to standard form is given in the following table.

Types of

constraint

Extra variable

needed

Coefficient of extra

variables In the

objective function

Presence of extra

variable in initial

solution mix

Max Z Min Z

Less than or

equal to ≤

A slack variable is

added

0 0 Yes

Greater than

or equal to (≥)

A surplus variable

is subtracted and

artificial variable is

added

0 0 No

-M +M Yes

Equal to (=) Only an artificial

variable is added

-M +M Yes

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Simplex cont…

A slack variable represents unused resource, either in the form of time on a machine, labor hour, money, warehouse space or any number of such resources in various business problems.

Since variable yields no profit, therefore such variable are added to the original objective function with zero coefficients.

A surplus variable represents amount by which solution values exceed a resource. These variables are also called negative slack variables.

Surplus variables carry a zero coefficient in the objective function.

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Simplex cont…The major steps and activities to solve LP model by using simplex

method

If we get more variables & less equation, we can set extra variables equal to zero, to obtain a system of equal variables & equal equations. Such solution is called basic solution.

The variables having positive values in a basic feasible solution are called basic variable while the variables which are set equal to zero, so as to define a corner point are called non-basic variables.

Slack variables are the fictitious variables which indicate how much of a particular resource remains unused in any solution

Column denotes the unit contribution margin.

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Row denotes the contribution margin lost if one unit is brought into the solution

•

Row denotes the Net Potential contribution or the Net unit margin potential, per unit.

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The rules used under simplex method, for solving a linear programming problem are as follows:-

1. Convert the given problem into Standard maximization Problem

2. Convert all ≤ constraints to equalities by adding a different slack variable for each one of them.

3. Construct the initial simplex tableau with all slack variables in the Basic Variables. The last row in the table contains the coefficient of the objective function (row).

4. Determine whether the current tableau is optimal. That is: If all the right hand side values are non-negative (called the feasibility condition). If all elements of the last row, that is rows are non-positive (called, the optimality condition). The current tableau contains an optimal solution. Otherwise, go to the next step.

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5. If the current BASIC VRIABLES is not optimal, determine, which non basic variable should become a basic variable and, which basic variable should become a non basic variable.

To find the new BASIC VRIABLES with the better objective function value, perform the following tasks:

Identify the entering variable: The entering variable is the one with the largest positive value (In case of a tie, select the variable that corresponds to the left most of the columns).

Identify the outgoing variable: The outgoing variable is the one with smallest non-negative column ratio (to find the column ratios, divide the RHS column by the entering variable column, wherever possible). In case of a tie select the variable that corresponds to the upper most of the tied rows.

Generate the new tableau

Select the largest value of row. The column, under which this value falls, is the pivot-column.

Pivot-row selection rule. Find the ratio of quantity to the corresponding pivot-column coefficient. The pivot-row selected is the variable having the least ratio

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Remark: Rows having negative or zero coefficients in the pivot-column are to be neglected. The coefficient, which is in both, the pivot-row & the pivot column, is called the pivot-element or pivot-number. Up-dating Pivot-row. Pivot-row, also called replaced rows, are updated as under all elements of old-row divided by Pivot-element. Now, in the basic activities column, write the pivot-column variable in place of the pivot-row variable. i.e.; the pivot-row variable is to be replaced by the pivot-column variable.

Up-Dating all other rows. Update all other rows by updating the formulae.

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Up-dating rows. Each is obtained as the sum of the products of the column coefficients multiplied by the corresponding coefficient in the jth column. (i.e.) the Quantity column). It is then subtracted from row values to get values. This pivoting is to be repeated till no positive coefficients exist in the row, the optimal solution is known.

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Simplex algorism (Maximization Case)

A Standard Maximization Problem is the one that satisfies the following four conditions

1. The Objective function is to be maximized.

2. All the inequalities are of ≤ type.

3. All right hand constants are non-negative.

4. All variables are non-negative.

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Example 1.Maximize Z = 40X1+35X2

Subject to:

2X1+3X2≤60 raw material constraint

4X1+3X2≤96 Labor hour constraint

X1&X2≥0

Solution:

Step one standardization of the problem

Maximize Z = 40X1+35X2+0S1+0S2

Subject to:

2X1+3X2+S1=60 raw material constraint

4X1+3X2+S2=96 Labor hour constraint

X1,X2,S1,S2≥0

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Step two obtaining the initial tableau and solution

Test for optimality right hand side should be +ve and the Cj-Zj row should be –ve. So the solution is not optimal.

BVCj 40 35 0 0 RHSV

X1 X2 S1 S2 bi

S1 0 2 3 1 0 60

S2 0 4 3 0 1 96

Zj 0 0 0 0

Cj-Zj 40 35 0 0

Constant values

Bas

ic

vari

able

s

Contribution margin

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Step three identify the entering and leaving variable

4* is pivot point

BVCj 40 35 0 0 RH

SVBi/EV

X1 X2 S1 S2 bi

S1 0 2 3 1 0 60 30

S2 0 4* 3 0 1 96 24

Zj 0 0 0 0

Cj-Zj 40 35 0 0

Leaving variable

Entering variable (EV)

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Simplex table two

The solution is not optimal because we have +ve value in Cj-Zj row

BVCj 40 35 0 0 RH

SV

X1 X2 S1 S2 bi

S1 0 0 3/2 1 -1/2 12

X1 40 1 3/4 0 1/4 24

Zj 40 30 0 10 960

Cj-Zj 0 5 0 -10

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3/2* pivot element

BVCj 40 35 0 0 RH

SVBi/EV

X1 X2 S1 S2 bi

S1 0 0 3/2* 1 -1/2 12 8

X1 40 1 3/4 0 1/4 24 32

Zj 40 30 0 10 960

Cj-Zj 0 5 0 -10

Entering variable (EV)

Leaving variable

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Simplex table 3

The solution is optimal because all the values of Cj-Zj row is negative and zero and all the right hand side values are positive

BVCj 40 35 0 0 RHS

V

X1 X2 S1 S2 bi

X2 35 0 1 2/3 -1/3 8

X1 40 1 0 -1/2 1/2 18

Zj 40 35 10/3 25/3 1000

Cj-Zj 0 0 -10/3 -25/3